The smallest signless Laplacian spectral radius of graphs with a given clique number

Linear Algebra and its Applications 439 (2013) 2562–2576

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Linear Algebra and its Applications www.elsevier.com/locate/laa

The smallest signless Laplacian spectral radius of graphs with a given clique number ✩ Jing-Ming Zhang a,b , Ting-Zhu Huang a , Ji-Ming Guo c,∗ a

School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 611731, China b College of Science, China University of Petroleum, Shandong, QingDao, 266580, China c College of Science, East China University of Science and Technology, Shanghai, 200237, China

a r t i c l e

i n f o

Article history: Received 10 January 2013 Accepted 22 July 2013 Available online 26 August 2013 Submitted by R. Brualdi MSC: 05C50

a b s t r a c t The signless Laplacian spectral radius of a graph G is the largest eigenvalue of its signless Laplacian matrix. In this paper, the first four smallest values of the signless Laplacian spectral radius among all connected graphs with maximum clique of size greater than or equal to 2 are obtained. © 2013 Elsevier Inc. All rights reserved.

Keywords: Signless Laplacian spectral radius Clique number Signless Laplacian characteristic polynomial

1. Introduction Let G = ( V , E ) be a simple connected graph with vertex set V = { v 1 , v 2 , . . . , v n } and edge set E. Its adjacency matrix A (G ) = (ai j ) is defined as n × n matrix (ai j ), where ai j = 1 if v i is adjacent to v j ; and ai j = 0, otherwise. Denote by d( v i ) or d G ( v i ) the degree of the vertex v i . Let Q (G ) = D (G ) + A (G ) be the signless Laplacian matrix of graph G, where D (G ) = diag(d( v 1 ), d( v 2 ), . . . , d( v n )) denotes the diagonal matrix of vertex degrees of G. It is well known that A (G ) is a real symmetric matrix and Q (G ) is a positive semidefinite matrix. The eigenvalues of Q (G ) can be ordered as

q1 (G )  q2 (G )  · · ·  qn (G )  0, ✩ This research is supported by NSFC (Nos. 11371372, 61170311), Sichuan Province Sci. & Tech. Research Project (No. 12ZC1802). Corresponding author. E-mail addresses: [email protected] (J.-M. Zhang), [email protected] (T.-Z. Huang), [email protected] (J.-M. Guo).

*

0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.07.027

J.-M. Zhang et al. / Linear Algebra and its Applications 439 (2013) 2562–2576

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Fig. 1. Kite graph P K n−ω,ω .

respectively. The largest eigenvalues of A (G ) and Q (G ) are called the spectral radius and the signless Laplacian spectral radius of G, denoted by ρ (G ) and q(G ), respectively. It is easy to see that if G is connected, then Q (G ) is nonnegative irreducible matrix. By the Perron–Frobenius theory, q(G ) has multiplicity one and there exists a unique positive unit eigenvector corresponding to q(G ). We refer to such an eigenvector corresponding to q(G ) as the Perron vector of G. Denote by P n , C n and K 1,n−1 the path, the cycle and the star on n vertices, respectively. The characteristic polynomial of a square matrix B is denoted by Φ( B ) = det(xI − B ). In particular, if B = Q (G ), we write Φ( Q (G )) by Ψ (G ). Let X be an eigenvector of G corresponding to q(G ). It will be convenient to associate with X a labeling of G in which vertex v i is labeled xi (or x v i ). Such labelings are sometimes called “valuation” [1]. The investigation on the spectral radius of graphs is an important topic in the theory of graph spectra. The recent developments on this topic also involve the problem concerning graphs with maximal or minimal spectral radius, Laplacian spectral radius, of a given class of graphs, respectively. In [2], the first three smallest values of the Laplacian spectral radii among all connected graphs with maximum clique size ω are given. And in [3], it is shown that among all connected graphs with maximum clique size ω , the minimum value of the spectral radius is attained for a kite graph P K n−ω,ω , where P K n−ω,ω is a graph on n vertices obtained from the path P n−ω and the complete graph K ω by adding an edge between an end vertex of P n−ω and a vertex of K ω (shown in Fig. 1). In this paper, the first four smallest values of the signless Laplacian spectral radius are obtained among all connected graphs with maximum clique size ω . Let n,ω be the set of all connected graphs of order n with a maximum clique size ω (2  ω  n). It is easy to see that ω,ω = { K ω }. By a direct calculation, we have q( K ω ) = 2ω − 2. If G ∈ ω+1,ω , then from the Perron–Frobenius theory, the first ω − 1 smallest values of the signless Laplacian spectral radius of ω+1,ω are P K 1,ω;i (0  i  ω − 2), respectively, where P K 1,ω;i is the graph obtained from P K 1,ω by adding i (0  i  ω − 2) edges. So in the following, we consider that n  ω + 2. 2. Preliminaries Let G be a graph and G − e be the graph obtained from G by deleting an edge e of G. In order to complete the proof of our main result, we need the following lemmas. Lemma 2.1. (See [4].) Let e be an edge of the graph G. Then

q1 (G )  q1 (G − e )  q2 (G )  q2 (G − e )  · · ·  qn (G )  qn (G − e )  0. In particular, for the signless Laplacian spectral radius of a graph, by the well-known Perron– Frobenius theory, we have Lemma 2.2. Let G be a connected graph and H be a proper subgraph of G. Then q( H ) < q(G ). Lemma 2.3. (See [5].) Let G be a graph on n vertices with vertex degrees d( v 1 ), d( v 2 ), . . . , d( v n ). Then

q(G ) 

max

v i v j ∈ E (G )





d( v i ) + d( v j ) .

For a connected graph G, equality holds if and only if G is regular or semi-regular bipartite.

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Fig. 2. Grafting an edge.

An internal path of a graph G is a sequence of vertices v 1 , v 2 , . . . , v k with k  2 such that: (1) The vertices in the sequence are distinct (except possibly v 1 = v k ); (2) v i is adjacent to v i +1 (i = 1, 2, . . . , k − 1); (3) The vertex degrees d( v i ) satisfy d( v 1 )  3, d( v 2 ) = · · · = d( v k−1 ) = 2 (unless k = 2) and d( v k )  3. Lemma 2.4. (See [6].) Let P : v 1 v 2 · · · v k (k  2) be an internal path of a connected graph G. Let G  be a graph obtained from G by subdividing some edge of P . Then we have q(G  ) < q(G ). Let v be a vertex of a graph G and suppose that the paths v k+1 v k · · · v 2 v 1 and ul+1 ul · · · u 2 u 1 are identified to the vertex v of G. Observe that in the new graph G k,l , v = v k+1 = ul+1 (shown in Fig. 2). Let G + e be the graph obtained from G by adding a new edge e and

G k+1,l−1 = G k,l − u 1 u 2 + v 1 u 1 . We call that G k+1,l−1 is obtained from G k,l by grafting an edge (see Fig. 2). Lemma 2.5. (See [6].) Let G be a connected graph on n  2 vertices and v is a vertex of G. Let G k,l and G k+1,l−1 (k  l  1) be the graphs defined as above. Then q(G k,l ) > q(G k+1,l−1 ). Let Q v (G ) be the principal submatrix of Q (G ), formed by deleting the row and column corresponding to the vertex v. Similarly, if H is a subgraph of G, let Q H (G ) be the principal submatrix of Q (G ), formed by deleting the rows and columns corresponding to all the vertices of H in V ( H ). Lemma 2.6. (See [7].) Let v be a vertex of a graph G, let ϕ ( v ) be the set of all cycles containing v. Then the signless Laplacian characteristic polynomial Ψ (G ) satisfies

          Ψ ( G ) = x − d( v ) Φ Q v ( G ) − Φ Q v w (G ) − 2 Φ Q Z (G ) , w v∈E

Z ∈ϕ ( v )

where the first summation extends over those vertices w adjacent to v, and the second summation extends over all Z ∈ ϕ ( v ). Let v be a vertex of the graph G and N ( v ) be the set of vertices adjacent to v. Lemma 2.7. (See [5,8].) Let G be a connected graph, and u , v be two vertices of G. Suppose that v 1 , v 2 , . . . , v s ∈ N ( v )\{ N (u ) ∪ {u }} (1  s  d( v )) and x = (x1 , x2 , . . . , xn ) is the Perron vector of G, where xi corresponds to the vertex v i (1  i  n). Let G ∗ be the graph obtained from G by deleting the edges v v i and adding the edges uv i (1  i  s). If xu  x v , then q(G ) < q(G ∗ ). Let S (G ) be the subdivision graph of G obtained by subdividing every edge of G. Lemma 2.8. (See [9,10].) Let G be a graph on n vertices and m edges, Φ(G ) = det(xI − A (G )), Ψ (G ) = det(xI − Q (G )). Then Φ( S (G ), x) = xm−n Ψ (G , x2 ).

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Fig. 3. H 1 and H 2 .

Lemma 2.9. (See [11].) Let v be a vertex of G, and let ϕ ( v ) be the collection of circuits containing v, and let V ( Z ) denote the set of vertices in the circuit Z . Then the characteristic polynomial Φ(G ) satisfies

Φ(G ) = xΦ(G − v ) −



Φ(G − v − w ) − 2

w v∈E



  Φ G − V (Z) ,

Z ∈ϕ ( v )

where the first summation extends over those vertices w adjacent to v, and the second summation extends over all Z ∈ ϕ ( v ). Let W n be the tree on n vertices obtained from P n−4 by attaching two new pendant edges to each end vertex of P n−4 , respectively. Lemma 2.10. (See [12].) Suppose that G = W n is a connected graph and uv is an edge on an internal path of G. Let G uv be the graph obtained from G by subdividing the edge uv. Then ρ (G uv ) < ρ (G ). 3. Main results Let H 1 be the graph obtained from K ω and a path P 4 by joining some vertex of K ω and some non-pendant vertex of P 4 by a path with length 2; and let H 2 be the graph obtained from K ω by attaching two pendant edges at two different vertices of K ω (see Fig. 3). Lemma 3.1. Let H 1 and H 2 be the graphs as above (see Fig. 3). If ω  3, then q( H 2 ) > q( H 1 ). Proof. For 11  ω  3, we have q( H 2 ) > q( H 1 ). In the following, we suppose that Lemma 2.6, we have

ω  12. From

      Ψ ( H 1 ) = (x − 2)Φ Q v 5 ( H 1 ) − Φ Q v 5 v 6 ( H 1 ) − Φ Q v 5 v 4 ( H 1 )    = (x − 2)(x − ω + 2)ω−2 x4 − 7x3 + 14x2 − 8x + 1 (x − 2ω + 2)(x − ω + 1) − 1   − (x − 2ω + 3)(x − ω + 2)ω−2 x4 − 7x3 + 14x2 − 8x + 1    − (x − ω + 2)ω−2 (x − 2ω + 2)(x − ω + 1) − 1 x3 − 4x2 + 4x − 1      = (x − ω + 2)ω−2 x7 − (3ω + 6)x6 + 2ω2 + 23ω x5 − 18ω2 + 43ω − 44 x4       + 54ω2 − 26ω − 49 x3 − 64ω2 − 117ω + 28 x2 + 26ω2 − 65ω + 33 x   − 2ω 2 − 6ω + 4 = (x − ω + 2)ω−2 g 1 (x). By a direct calculation, we have

     Ψ ( H 2 ) = (x − ω + 2)ω−3 x5 − (4ω − 2)x4 + 5ω2 − 2ω − 6 x3 − 2ω3 + 4ω2 − 18ω + 8 x2     + 4ω3 − 11ω2 + 2ω + 8 x − 2ω3 − 10ω2 + 16ω − 8 = (x − ω + 2)ω−3 g 2 (x).

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Fig. 4. Graph P K ni −ω,ω , where i = ω + 1, . . . , n − 1.

n−2 Fig. 5. Graph P K n−ω,ω .

From Lemmas 2.2 and 2.3, we have 2ω − 1 > q( H 1 ) > q( K ω ). For ω  12, we have q( H 1 − v 5 v 6 ) = q( K ω ). From Lemma 2.1, we have q( H 1 )  q( H 1 − v 5 v 6 )  q2 ( H 1 ). So, 2ω − 1 > q( H 1 ) > q( K ω )  q2 ( H 1 ). From Lemmas 2.2 and 2.3, we have 2ω > q( H 2 ) > q( K ω ). Thus q( H 1 ) and q( H 2 ) are the largest roots of the equations g 1 (x) = 0 and g 2 (x) = 0, respectively. By a direct calculation, we have for ω  12,



g 1 2ω − 2 +

g 1 2ω − 2 +

g 2 2ω − 2 + Thus for

1



= − 48ω4 + 440ω3 − 1728ω2 + 3948ω +

ω

2



2

2276

ω

3

−

777

ω

4

+

167

ω

5

−

20

ω

6

+

1

ω7

ω

−

+



28264

ω3

−

15328

ω4

= −12ω2 + 54ω +

ω  12, we have q( H 2 ) > 2ω − 2 +

+

240

ω

5696

ω5 −

−

288

ω2

1280

+

ω6 240

ω3

4488

ω2

− 5931 < 0;

= 32ω5 − 384ω4 + 2096ω3 − 7080ω2 + 16714ω +

ω

ω

+

6158

+ −

128

ω7 128

ω4

38150

ω

−

37928

ω2

− 29080 > 0; +

2 ω > q( H 1 ). The result follows.

32

ω5

− 140 < 0.

2

Let P K ni −ω,ω be the graph obtained from the kite graph P K n−ω−1,ω (see Fig. 1) and an isolated

2 vertex v n by adding an edge v n v i (ω + 1  i  n − 1) (see Fig. 4). It is easy to see that P K 5ω,+ ω = H1

−1 and P K nn− ω,ω = P K n−ω,ω . n−2

−2 P K n−ω,ω = P K nn− Let ω,ω + v n−1 v n (see Fig. 5). Let G 0 be the graph obtained from S ( P K n−8,3 ) and an isolated vertex, say u, by adding an edge −3 n−4 between the pendant vertex and u. Let G 1 = P K nn− 3,3 + v n−1 v n , G 2 = P K n−3,3 + v n−1 v n and C n−1,1 be the graph obtained from one cycle C n−1 and an isolated vertex by adding an edge between the isolated vertex and one vertex of C n−1 (see Fig. 6).

Lemma 3.2. For n  6, let P K ni −ω,ω and C n−1,1 be the graphs defined as above (see Figs. 4 and 6). Then









−2 n −3 q( P n ) < q(C n ) < q P K nn− 2,2 < q P K n−2,2 < q (C n−1,1 ). −3 Proof. Note that P K nn− 2,2 is a proper subgraph of C n−1,1 . From Lemma 2.2, we have q( P n ) < q (C n ) n−3 −2 n−3 and q( P K n−2,2 ) < q(C n−1,1 ). Note that P K nn− 2,2 can be obtained from P K n−2,2 by the grafting an n−2 n−3 −2 edge. From Lemma 2.5, we have q( P K n−2,2 ) < q( P K n−2,2 ) (n  6). For n  5, we have that P K nn− 2, 2

J.-M. Zhang et al. / Linear Algebra and its Applications 439 (2013) 2562–2576

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Fig. 6. Graphs G 1 , G 2 , C n−1,1 .

contains P K 33,2 as a subgraph. By a direct calculation, we have q( P K 33,2 ) ≈ 4.16. It is easy to prove

−2 that q(C n ) = 4. Thus, from Lemma 2.2, we have q(C n ) < q( P K nn− 2,2 ). The result follows.

2

Theorem 3.3. Among all connected graphs on n vertices with maximum clique size ω = 2 and n  6, the first −2 n−3 four smallest signless Laplacian spectral radii are exactly obtained for P n , C n , P K nn− 2,2 , P K n−2,2 , respectively. Proof. Let G be a connected graph with maximum clique size ω = 2 and n  6 vertices. From −2 n−3 Lemma 3.2, we have q( P n ) < q(C n ) < q( P K nn− 2,2 ) < q ( P K n−2,2 ). Thus, we only need to prove that

−3 n−2 n−3 q(G ) > q( P K nn− 2,2 ) if G = P n , C n , P K n−2,2 , P K n−2,2 . Suppose that u ∈ V ( G ) and d(u ) = max{d( v ) | v ∈ V (G )}. If d(u )  4, then G contains K 1,4 as a subgraph. By a direct calculation, we have q( K 1,4 ) = −3 5 > q(C 4,1 ) ≈ 4.48119. Thus from Lemmas 2.2, 2.4 and 3.2, we have q(G ) > q(C n−1,1 ) > q( P K nn− 2, 2 ) . Suppose that d(u )  3 and G contains a cycle. Note that G = C n . Thus, we have d(u ) = 3. So, G contains C m−1,1 (m  n) as a subgraph. From Lemmas 2.2 and 2.4, we have q(C n−1,1 )  q(C m−1,1 )  q(G ). −3 n−2 n−3 Hence, q(G ) > q( P K nn− 2,2 ). Suppose that d(u )  3 and G is a tree. Note that G = P n , P K n−2,2 , P K n−2,2 . −3 From Lemma 2.5, we have q(G ) > q( P K nn− 2,2 ). The result follows.

2

Lemma 3.4. For n  8, let P K ni −ω,ω and G 1 be the graphs defined as above (see Figs. 4 and 6). Then





−4 q(G 1 ) < q P K nn− 3,3 . −4 Proof. For 8  n  12, by a direct calculation, we have q(G 1 ) < q( P K nn− 3,3 ). If n  13, from Lem-

−4 9 mas 2.2 and 2.10, we have 2.15846 < ρ ( S ( P K 10,3 )) < ρ ( S ( P K nn− 3,3 )) < ρ ( S ( P K 10,3 )) < 2.15851. Similarly, we have 2.15846 < ρ ( S (G 1 )) < 2.15851. From Lemma 2.8, we only need to prove that ρ ( S ( P K nn−−34,3 )) > ρ ( S (G 1 )). From Lemma 2.9, we have

    −4  = x9 − 8x7 + 21x5 − 20x3 + 5x Φ(G 0 ) Φ S P K nn− 3,3   − x8 − 6x6 + 11x4 − 7x2 + 1 Φ(G 0 − u ) = f 1 (x)Φ(G 0 ) − f 2 (x)Φ(G 0 − u ).    10 Φ S (G 1 ) = x − 10x8 + 34x6 − 46x4 + 20x2 Φ(G 0 )   − x9 − 9x7 + 26x5 − 26x3 + 4x Φ(G 0 − u ) 

= f 3 (x)Φ(G 0 ) − f 4 (x)Φ(G 0 − u ).

(3.1)

For 2.15846 < x < 2.15851, we have

f 1 (x) > 2.158469 − 8 × 2.158517 + 21 × 2.158465 − 20 × 2.158513 + 5 × 2.15846 > 0, f 3 (x) > 2.1584610 − 10 × 2.158518 + 34 × 2.158466 − 46 × 2.158514 + 20 × 2.158462 > 0. By a direct calculation, we have

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−4 f 1 (x)Φ S (G 1 ) − f 3 (x)Φ S P K nn− 3,3



  = f 2 (x) f 3 (x) − f 1 (x) f 4 (x) Φ(G 0 − u )   = x16 − 14x14 + 76x12 − 202x10 + 269x8 − 158x6 + 24x4 Φ(G 0 − u ) = F 1 (x)Φ(G 0 − u ).

For 2.15846 < x < 2.15851, we have

F 1 (x) = 16x15 − 196x13 + 912x11 − 2020x9 + 2152x7 − 948x5 + 96x3

> 16 × 2.1584615 − 196 × 2.1585113 + 912 × 2.1584611 − 2020 × 2.158519 + 2152 × 2.158467 − 948 × 2.158515 + 96 × 2.158463 ≈ 10288 > 0. Since F 1 (2.15846) ≈ 47 > 0, we have, F 1 (x) > 0 for 2.15846 < x < 2.15851. Note that





ρ (G 0 − u ) < ρ S (G 1 )

and





2.15846 < ρ S (G 1 ) < 2.15851.

ρ ( S (G 1 ))  x < 2.15851, we have Φ(G 0 − u ) > 0. Then         −4  f 1 (x)Φ S (G 1 ) > f 3 (x)Φ S P K nn− x ∈ ρ S (G 1 ) , 2.15851 . 3,3 ,

Hence, for

Thus,

ρ ( S (G 1 )) < ρ ( S ( P K nn−−34,3 )). 2 n−2

Lemma 3.5. For n  11, let P K ni −ω,ω and P K n−ω,ω be the graphs defined as above (see Figs. 4, 5). Then









−4 n−2 q P K nn− 3,3 < q P K n−3,3 . n−2

−4 Proof. If 11  n  15, by a direct calculation, we have q( P K nn− 3,3 ) < q ( P K n−3,3 ). If n  16, by a similar reasoning as that of Lemma 3.4, we have

 

−4 2.15846 < ρ S P K nn− 3,3



  n−2  < 2.15851 and 2.15846 < ρ S P K n−3,3 < 2.15851.

From Lemma 2.8, we only need to prove that have

n−2

ρ ( S ( P K nn−−34,3 )) < ρ ( S ( P K n−3,3 )). From Lemma 2.9, we

   n−2   10 Φ S P K n−3,3 = x − 10x8 + 34x6 − 48x4 + 27x2 − 4 Φ(G 0 )   − x9 − 9x7 + 26x5 − 29x3 + 11x Φ(G 0 − u ) = f 5 (x)Φ(G 0 ) − f 6 (x)Φ(G 0 − u ). For 2.15846 < x < 2.15851, we have

f 5 (x) > 2.1584610 − 10 × 2.158518 + 34 × 2.158466 − 48 × 2.158514 + 27 × 2.158462 − 4

≈ 0.8 > 0. Combined the above equation with Eq. (3.1), we have

 

  n−2  − f 1 (x)Φ S P K n−3,3   = f 1 (x) f 6 (x) − f 2 (x) f 5 (x) Φ(G 0 − u )   = −x16 + 14x14 − 77x12 + 214x10 − 320x8 + 250x6 − 84x4 + 4 Φ(G 0 − u )

−4 f 5 (x)Φ S P K nn− 3,3



= F 2 (x)Φ(G 0 − u ). Since

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F 2 (x) = −16x15 + 196x13 − 924x11 + 2140x9 − 2560x7 + 1500x5 − 336x3

< −16 × 2.1584615 + 196 × 2.1585113 − 924 × 2.1584611 + 2140 × 2.158519 − 2560 × 2.158467 + 1500 × 2.158515 − 336 × 2.158463 ≈ −10690 < 0 and F 2 (2.15851) ≈ 42 > 0, we have F 2 (x) > 0 for 2.15846 < x < 2.15851. From the proof of Lemma 3.4, we have Φ(G 0 − u ) > 0. Then

 

4 f 5 (x)Φ S P K nn− −3,3

Thus,



  n−2  > f 1 (x)Φ S P K n−3,3 ,

x∈

  





ρ S P K nn−−34,3 , 2.15851 .

n−2

ρ ( S ( P K nn−−34,3 )) < ρ ( S ( P K n−3,3 )). 2

n−2 Lemma 3.6. For ω  4, let P K ni −ω,ω and P K n−ω,ω be the graphs defined as above (see Figs. 4 and 5). Then −3 n−2 q( P K nn− ω,ω ) > q( P K n−ω,ω ). n−2 Proof. Let X = (x1 , x2 , . . . , xn ) T be the Perron vector of P K n−ω,ω , where xi corresponds to v i . It is n−2 easy to prove that xn = xn−1 . From ( D + A ) X = q( P K n−ω,ω ) X , we have

⎧   n −2 ⎨q P K n−ω,ω xn = 3xn + xn−2 ,  n −2 ⎩ q  P K n−ω,ω xn−2 = 3xn−2 + 2xn + xn−3 . n−2

From Lemma 2.1, we have q( P K n−ω,ω )  q( K ω ) = 2ω − 2  6 for tions, we have



ω  4. Thus, from the above equa-

xn−2  3xn , xn−3  3xn−2 − 2xn .

Thus, we have







−3 n−2 q P K nn− ω,ω − q P K n−ω,ω



  n −2  −3  T  X T Q P K nn− ω,ω X − X Q P K n−ω,ω X = xn2−3 − xn2−2 − 4xn2 + 2xn xn−3 − 2xn xn−2  (3xn−2 − 2xn )2 − 4xn2 + 2xn (3xn−2 − 2xn ) − 2xn xn−2 − xn2−2 = 8xn−2 (xn−2 − xn ) − 4xn2  16xn−2 xn − 4xn2 > 0. n−2

−3 So, q( P K nn− ω,ω ) > q( P K n−ω,ω ).

2

Let H 3 be the graph obtained from K ω by attaching two pendant edges at some vertex of K ω ; let H 4 be the graph obtained from K ω and P 2 by adding two edges between two vertices of K ω and two end vertices of P 2 (see Fig. 7). Theorem 3.7. Among all connected graphs on n vertices with maximum clique size ω = 3 and n  13, the −2 n−3 first four smallest signless Laplacian spectral radii are exactly obtained for P K n−3,3 , P K nn− 3,3 , P K n−3,3 , G 1 , respectively. Proof. Let G be a connected graph with maximum clique size ω = 3 and n  13 vertices. Suppose that K 3 is a maximum clique of G. From Lemmas 2.2, 2.4 and 2.5, we have









−3 n −2 q(G 1 ) > q P K nn− 3,3 > q P K n−3,3 > q ( P K n−3,3 ). −2 n−3 Thus, we only need to prove that q(G ) > q(G 1 ) if G = P K n−3,3 , P K nn− 3,3 , P K n−3,3 , G 1 .

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Fig. 7. Graphs H 3 , H 4 , F g .

2 3 Fig. 8. Graphs M ω , Mω , H5.

For n  13, from the proof of Lemma 3.4, we have 2.15846 < ρ ( S (G 1 )) < 2.15851. We distinguish the following three cases. Case 1. If there exist at least two vertices outside of K 3 that are adjacent to some vertices of K 3 , then we have that G contains either H 2 or H 3 as a subgraph. If G contains H 2 as a subgraph, from Lemma 2.2, we have













ρ S (G ) > ρ S ( H 2 ) ≈ 2.22158 > 2.15851 > ρ S (G 1 ) . If G contains H 3 as a subgraph, from Lemma 2.2, we have













ρ S (G ) > ρ S ( H 3 ) ≈ 2.3 > 2.15851 > ρ S (G 1 ) . From Lemma 2.8, we have q(G ) > q(G 1 ). Case 2. Suppose that there exists a vertex, say u, which does not belong to K 3 , such that u is adjacent to at least two vertices of K 3 . From Lemmas 2.2 and 2.4, we have













ρ S (G ) > ρ S ( H 4 ) ≈ 2.26 > 2.15851 > ρ S (G 1 ) . Thus, q(G ) > q(G 1 ). Case 3. Suppose that there uniquely exists a vertex u, which does not belong to K 3 , such that u is adjacent to a vertex of K 3 . Suppose that G − V ( K 3 ) is a tree. If there exists some vertex, say v, of G − V ( K 3 ) such that d( v )  4, then G contains K 1,4 as a subgraph. We have ρ ( S ( K 1,4 )) ≈ 2.23607 > 2.15851 > ρ ( S (G 1 )). From Lemmas 2.2 and 2.8, we have q(G ) > q(G 1 ); otherwise, note that G = −2 n−3 n−4 P K n−3,3 , P K nn− 3,3 , P K n−3,3 , then from Lemmas 2.2, 2.4, 2.5 and 3.4, we have q ( G )  q ( P K n−3,3 ) > q ( G 1 ). If G − V ( K 3 ) contains one triangle as a subgraph, then from Lemmas 2.2, 2.4, 3.4 and 3.5, we n−2

−4 have q(G )  q( P K n−3,3 ) > q( P K nn− 3,3 ) > q ( G 1 ). If G − V ( K 3 ) contains C 4 as a subgraph, note that G = G 1 , then from Lemmas 2.2 and 2.4, we have q(G ) > q(G 1 ). If G − V ( K 3 ) contains C g ( g  5) as a subgraph, then from Lemma 2.1, we can construct a graph F g by deleting edges and isolated vertices (if exist) such that q(G )  q( F g ), where F g is the graph obtained from K 3 and a cycle C g by joining a vertex of K 3 and a vertex of C g with a path with length k + 1 and | V ( F g )|  n (see Fig. 7). Then from −4 Lemmas 2.2, 2.4 and 3.4, we have q(G )  q( F g ) > q( P K nn− 3, 3 ) > q ( G 1 ) .

2

i Let M ω (ω  3, 2  i  ω − 1) be the graph obtained from K ω and P 2 by adding i edges between an end vertex of P 2 and i vertices of K ω , H 5 be the graph obtained from K ω and two isolated vertices by adding three edges between two isolated vertices and three vertices of K ω (see Fig. 8).

J.-M. Zhang et al. / Linear Algebra and its Applications 439 (2013) 2562–2576

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Theorem 3.8. Among all connected graphs on n vertices with maximum clique size ω  4 and n  ω + 5, the first four smallest signless Laplacian spectral radii are exactly obtained for P K n−ω,ω , P K nn−−ω2 ,ω , n−2 −3 P K n−ω,ω , P K nn− ω,ω , respectively.

Proof. Let G be a connected graph with maximum clique size ω  4 and n  ω + 5 vertices. Suppose that K ω is a maximum clique of G. From Lemmas 2.2, 2.5 and 3.6, we have













−3 n −2 n−2 q P K nn− ω,ω > q P K n−ω,ω > q P K n−ω,ω > q( P K n−ω,ω ). n−2

−3 n−2 Thus, we only need to prove that q(G ) > q( P K nn− ω,ω ) if G = P K n−ω,ω , P K n−ω,ω , P K n−ω,ω and

3 P K nn− −ω,ω . We distinguish the following three cases. Case 1. Suppose that there exist at least two vertices outside of K ω that are adjacent to some vertices of K ω . Then we have that G contains either H 2 or H 3 as a proper subgraph. If G contains H 2 as a proper subgraph, from Lemmas 2.2, 2.4 and 3.1, we have





−3 q(G ) > q( H 2 ) > q( H 1 )  q P K nn− ω,ω .

(3.2)

If G contains H 3 as a proper subgraph, then from Lemmas 2.2, 2.7 and Eq. (3.2), we have





−3 q(G ) > q( H 3 ) > q( H 2 ) > q P K nn− ω,ω .

Case 2. Suppose that there exists a vertex, say u, which does not belong to K ω , such that u is 2 adjacent to at least two vertices of K ω , then G contains M ω − v 1 as a subgraph, where v 1 is the 2 pendant vertex of M ω . From Lemmas 2.2, 2.4 and Eq. (3.2), we have









−3 2 q(G ) > q M ω − v 1 > q( H 4 ) > q( H 2 ) > q P K nn− ω,ω .

Case 3. Suppose that there uniquely exists a vertex u which does not belong to K ω such that u −2 n−3 is adjacent to a vertex of K ω . If G − V ( K ω ) is a tree, note that G = P K n−ω,ω , P K nn− ω,ω , P K n−ω,ω , n−2

−3 then from Lemmas 2.2, 2.4 and 2.5, we have q(G ) > q( P K nn− ω,ω ). If G = P K n−ω,ω and G − V ( K ω )

−3 contains C 3 or C 4 as a subgraph, then from Lemmas 2.2 and 2.4, we have q(G )  q(G ∗ ) > q( P K nn− ω ,ω ) , −3 where G ∗ is obtained from P K nn− ω,ω by adding an edge between two pendant vertices. If G − V ( K ω )

−4 contains C g ( g  5) as a subgraph, then from Lemmas 2.2 and 2.4, we have q(G ) > q( P K nn− ω ,ω ) >

−3 q( P K nn− ω ,ω ) .

2

2 2 Lemma 3.9. For ω  4, let M ω and H 5 be the graphs defined as above (see Fig. 8). Then q( H 5 ) > q( M ω ). 2 , where xi corresponds to v i . For ω = 4, 5, Proof. Let X = (x1 , x2 , . . . , xn ) T be the Perron vector of M ω 2 2 by a direct calculation, we have q( H 5 ) > q( M ω ). For ω  6, from ( D + A ) X = q( M ω ) X , we have

⎧   ⎪ q M 2 x = x1 + x2 , ⎪ ⎨  ω 1 2 q Mω x2 = 3x2 + 2x3 + x1 , ⎪ ⎪   ⎩ 2 q M ω x3 = ω x3 + (ω − 2)x4 + x2 + x3 .

Thus, we have

 





2 − 1 x1 , x2 = q M ω

x4 =

(3.3)

q ( M ω ) − ( w + 5)q ( M ω ) + (4w + 4)q( M ω ) − 2w x1 . 2( w − 2) 3

2

2

2

2

2 Note that 2ω > q( M ω )  q( K ω ) = 2ω − 2. From Eqs. (3.3) and (3.4), we have

(3.4)

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x4 x2

= > = >

2 2 2 q3 ( M ω ) − (ω + 5)q2 ( M ω ) + (4ω + 4)q( M ω ) − 2ω 2 ) − 1) 2(ω − 2)(q( M ω 2 2 2 2 q3 ( M ω ) − (ω + 5)q2 ( M ω ) + 2q2 ( M ω ) + 4q( M ω ) − 2ω 2 ) − 1) 2(ω − 2)(q( M ω 2 2 q ( M ω ) − (ω + 3)q2 ( M ω ) + 4q( M ω ) − 2ω 2 ) − 1) 2(ω − 2)(q( M ω 3

2

2 2 q3 ( M ω ) − (ω + 3)q2 ( M ω ) 2 ) − 1) 2(ω − 2)(q( M ω 2 2 q (Mω ) − (ω + 3)q( M ω ) 2(ω − 2) 2

>

 2 − (ω + 3) > 2ω − 2 − (ω + 3) = ω − 5  1. > q Mω 2 Then x4 > x2 . So, from Lemma 2.7, we have q( H 5 ) > q( M ω ).

2

2 2 Lemma 3.10. For ω  4, let M ω and H 3 be the graphs defined as above (see Figs. 7, 8). Then q( H 3 ) > q( M ω ). 2 ). From Lemmas 2.2 and 2.3, Proof. For 19  ω  4, by direct computations, we have q( H 3 ) > q( M ω 2 ) for we have 2ω > q( H 3 ) > q( K ω ) = 2ω − 2. In the following, we will prove that q( H 3 ) > q( M ω T ω  20. Let X = (x1 , x2 , . . . , xn ) be the Perron vector of H 3 , where xi corresponds to v i . From ( D + A ) X = q( H 3 ) X , we have

⎧ ⎨ q( H 3 )x1 = x1 + x2 , q( H 3 )x2 = (ω + 1)x2 + (ω − 1)xω + 2x1 , ⎩ q( H 3 )xω = (ω − 1)xω + (ω − 2)xω + x2 .

Then



q ( H 3 ) − 2ω + 3













q( H 3 ) − 1 q( H 3 ) − ω − 1 − 2 = (ω − 1) q( H 3 ) − 1 .

That is





q3 ( H 3 ) − (3ω − 1)q2 ( H 3 ) + 2ω2 + ω − 6 q( H 3 ) − 2ω2 + 6ω − 4 = 0. Let





r1 (x) = x3 − (3ω − 1)x2 + 2ω2 + ω − 6 x − 2ω2 + 6ω − 4. It is easy to prove that for 2ω > x  2ω − 2,





r1 (x) = 3x2 − 2(3ω − 1)x + 2ω2 + ω − 6 > 0. For

(3.5)

ω  20, we have r1 (2ω) = 4ω2 − 6ω − 4 > 0,



r 1 2ω − 2 +

2ω 2 + 5ω − 7

ω3

(3.6)

 =−

33

+

ω

+

735

ω8

99

ω2 −

−

343

ω9

115

ω3

−

135

ω4

+

563

ω5

−

540

ω6

−

231

ω7

< 0.

(3.7)

2 5ω−7 From Eqs. (3.5), (3.6), (3.7) and noting that 2ω > q( H 3 ) > 2ω − 2, we have q( H 3 ) > 2ω − 2 + 2ω + . ω3

ω  20, we have that q( M ω2 ) is the root of equation     r2 (x) = x4 − (3ω + 1)x3 + 2ω2 + 8ω − 12 x2 − 8ω2 − 14ω x + 4ω2 − 12ω + 8 = 0,

Similarly, for

x ∈ (2ω − 2, 2ω)

J.-M. Zhang et al. / Linear Algebra and its Applications 439 (2013) 2562–2576

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Fig. 9. Graphs H 6 , H 7 .

and for 2ω > x  2ω − 2, r2 (x) > 0. For



2



ω  20, we have

104

112

72

16

= −16ω − + 2 − 3 + 4 + 52 < 0, (3.8) ω ω ω ω ω

 2ω 2 + 5ω − 7 286 578 84 2104 4541 2758 5074 = 4ω + r 2 2ω − 2 + − 2 + 3+ − + + 3 4 5 6 ω ω ω ω ω ω ω ω7 r 2 2ω − 2 +

−

10729

ω8

+

5467

ω9

+

4606

ω10

−

6860

ω11

+

2401

ω12

− 76

> 0.

(3.9)

2 2 Thus, from Eqs. (3.8) and (3.9), we have q( M ω ) < 2ω − 2 + 2ω +ω53ω−7 . Hence, q( H 3 ) > q( M ω ). 2

2

Theorem 3.11. Let G be a graph on n vertices with maximum clique size ω  4 and n = ω + 2. Let P K 2,ω , H 2 , 2 H 4 and M ω be the graphs defined as above (see Figs. 1, 3, 7, 8). The first four smallest signless Laplacian 2 spectral radii are obtained for P K 2,ω , H 2 , H 4 , M ω , respectively. Proof. From Lemmas 2.2, 2.4 and 3.1, we have





2 > q( H 4 ) > q( H 2 ) > q( H 1 ) > q( P K 2,ω ). q Mω 2 2 Thus, we only need to prove that for G = P K 2,ω , H 2 , H 4 , M ω , q(G ) > q( M ω ). We distinguish the following two cases. Case 1. Suppose that there exists exactly one vertex outside of K ω that is adjacent to some vertices 2 3 of K ω . Note that G = P K 2,ω , M ω . Then G contains M ω (see Fig. 8) as a subgraph. From Lemma 2.2, 3 2 we have q(G )  q( M ω ) > q( M ω ). Case 2. Suppose that the two vertices outside of K ω that are all adjacent to some vertices of K ω . Note that G = H 2 , H 4 . Then G contains one of graphs H 3 and H 5 as a subgraph. From Lemma 3.10, 2 2 we have q( H 3 ) > q( M ω ). From Lemma 3.9, we have q( H 5 ) > q( M ω ). From Lemma 2.2, we have q(G ) > 2 q ( M ω ). 2

Let H 6 be the graph obtained from H 2 and an isolated vertex by adding an edge between a pendant vertex of H 2 and the isolated vertex, H 7 = H 6 − v 1 v 2 + v 1 v 4 (see Fig. 9). ω+1

P K 3,ω , H 2 and H 6 be the graphs defined as above (see Figs. 3, 5 and 9). Then Lemma 3.12. For ω  4, let





ω +1 q( H 6 ) > q( H 2 ) > q P K 3,ω . ω+1

Proof. For 8  ω  4, by direct computations, we have q( H 6 ) > q( H 2 ) > q( P K 3,ω ). In the following, we suppose that ω  9. From Lemma 2.6, we have

    ω +1    Ψ P K 3,ω = (x − ω + 2)ω−2 x5 − (3ω + 4)x4 + 2ω2 + 17ω − 8 x3 − 14ω2 + 9ω − 26 x2    + 26ω2 − 39ω + 1 x − 14ω2 + 34ω − 16 = (x − ω + 2)ω−2 g 3 (x).

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Fig. 10. Graphs H 8 , H 9 , H 10 , H 11 .

ω+1

ω+1

From Lemmas 2.1 and 2.3, we have 2ω − 1  q( P K 3,ω )  q( K ω ) = 2ω − 2  q2 ( P K 3,ω ). Thus ω+1

q ( P K 3,ω ) is the largest root of the equation g 3 (x) = 0. For

g 3 2ω − 2 +

g 3 2ω − 2 +

1

ω 2

 = −12ω2 + 72ω +

228



ω

−

= 8ω3 − 72ω2 + 274ω +

ω

173

ω2

966

ω

ω  9, we have

+

−

71

ω3

976

ω2

14

−

ω4

+

624

+

ω3

1

ω5

−

− 173 < 0;

224

ω4

+

32

ω5

− 632 > 0.

ω+1

2 Then q( P K 3,ω ) < 2ω − 2 + ω (ω  9). 2 From the proof of Lemma 3.1, we have q( H 2 ) > 2ω − 2 + ω (ω  5). Then we have for ω  9, ω + 1 2 q( H 2 ) > 2ω − 2 + ω > q ( P K 3,ω ). From Lemma 2.2, we have q( H 6 ) > q( H 2 ). The result follows. 2

Lemma 3.13. For ω  4, let H 6 and H 7 be the graphs defined as Fig. 9. Then q( H 7 ) > q( H 6 ). Proof. Let X = (x1 , x2 , . . . , xn ) T be the Perron vector of H 6 , where xi corresponds to v i . From ( D + A ) X = q( H 6 ) X , we have

⎧ q( H 6 )x1 = x1 + x2 , ⎪ ⎪ ⎨ q( H )x = 2x + x + x , 6 2 2 1 3 ⎪ q ( H ) x = ω x + ( ω − 2 )x4 + x2 + x5 , 6 3 3 ⎪ ⎩ q( H 6 )x4 = (ω − 1)x4 + (ω − 3)x4 + x3 + x5 .

Then we have







q( H 6 ) − ω + 2 x4 =



q( H 6 ) − ω + 1



q( H 6 ) − 2 −



1 q( H 6 ) − 1

 − 1 x2 .

From Lemma 2.2, we have q( H 6 ) > q( K ω ) = 2ω − 2. So, we have for





q( H 6 ) − ω + 2 < q( H 6 ) − ω + 1



q( H 6 ) − 2 −

1 q( H 6 ) − 1

ω  4 that  − 1.

Then, x4 > x2 . Thus, q( H 6 ) < q( H 6 − v 1 v 2 + v 1 v 4 ) = q( H 7 ) by Lemma 2.7.

2

Let H 8 be the graph obtained from H 3 and an isolated vertex by adding an edge between v ω and the isolated vertex; let H 9 be the graph obtained from H 3 and an isolated vertex by adding an edge between v 2 and the isolated vertex; let H 10 be the graph obtained from H 3 and an isolated vertex by adding an edge between one pendant vertex and the isolated vertex; let H 11 be the graph obtained 1 from P K 3ω,+ ω and an isolated vertex by adding an edge between v ω+1 and the isolated vertex (see Fig. 10). ω+1

1 Theorem 3.14. Let P K 3,ω , P K 3ω,+ ω , P K 3,ω and H 6 be the graphs defined as above (see Figs. 1, 4, 5 and 9). Among all connected graphs on n vertices with maximum clique size ω  4 and n = ω + 3, the first four

ω+1

1 smallest signless Laplacian spectral radii are obtained for P K 3,ω , P K 3ω,+ ω , P K 3,ω , H 6 , respectively.

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Proof. From Lemmas 2.2, 2.5 and 3.12, we have  ω +1    1 q( H 6 ) > q P K 3,ω > q P K 3ω,+ ω > q( P K 3,ω ). ω+1

1 Thus, we only need to prove that q(G ) > q( H 6 ) if G = P K 3,ω , P K 3ω,+ ω , P K 3,ω and H 6 . We distinguish the following four cases. Case 1. There exists exactly one vertex outside of K ω that is adjacent to only one vertex of K ω .

ω+1

1 Then G must be one of graphs P K 3,ω , P K 3ω,+ ω and P K 3,ω . Case 2. There exists one vertex outside of K ω that is adjacent to at least two vertices of K ω . 2 Then G contains M ω − v 1 (see Fig. 8) as a proper subgraph. From Lemmas 2.2 and 2.4, we have 2 q ( G ) > q ( M ω − v 1 ) > q ( H 6 ). Case 3. If there exist exactly two vertices outside of K ω that are adjacent to some vertices of K ω , then G contains H 6 or H 10 (see Fig. 10) as a subgraph. If G contains H 10 as a subgraph, then from Lemmas 2.2 and 2.7, we have q(G )  q( H 10 ) > q( H 6 ). If G contains H 6 as a subgraph, note that G = H 6 , then from Lemma 2.2, we have q(G ) > q( H 6 ). Case 4. If there exist three vertices outside of K ω that are adjacent to some vertices of K ω , then G contains one of graphs H 7 , H 8 and H 9 (see Figs. 9 and 10) as a subgraph. From Lemmas 2.7 and 3.13, we have q( H 9 ) > q( H 8 ) > q( H 7 ) > q( H 6 ). Then from Lemma 2.2, we have q(G ) > q( H 6 ). 2

Theorem 3.15. Among all connected graphs on n vertices with maximum clique size ω  4 and n = ω + 4, the

2 ω+2 ω+1 first four smallest signless Laplacian spectral radii are obtained for P K 4,ω , P K 4ω,+ ω , P K 4,ω and P K 4,ω (see Figs. 1, 4 and 5), respectively.

Proof. Let G be a connected graph with maximum clique size ω  4 and n = ω + 4 vertices. Suppose that K ω is a maximum clique of G. From Lemmas 2.2, 2.5 and 3.6, we have













1 ω +2 ω +2 q P K 4ω,+ ω > q P K 4,ω > q P K 4,ω > q( P K 4,ω ).

ω+2

1 ω+2 ω+1 Thus, we only need to prove that q(G ) > q( P K 4ω,+ ω ) if G = P K 4,ω , P K 4,ω , P K 4,ω and P K 4,ω . We distinguish the following three cases. Case 1. There exists exactly one vertex outside of K ω that is adjacent to one vertex of K ω . Subcase 1. Suppose that G − V ( K ω ) is a tree. If G contains exactly one pendant vertex, then G = 1 ω+2 P K 4,ω . If G contains exactly two pendant vertices, then G = P K 4ω,+ ω or G = P K 4,ω . If G contains

1 three pendant vertices, then G = H 11 (see Fig. 10). From Lemma 2.5, we have q( H 11 ) > q( P K 4ω,+ ω ). 1 ω+2 ω+1 Note that G = P K 4,ω , P K 4ω,+ ω and P K 4,ω . From Lemma 2.2, we have q(G ) > q( P K 4,ω ). Subcase 2. Suppose that G − V ( K ω ) contains a cycle. If G − V ( K ω ) contains C 4 , then G contains H 12 1 as a subgraph, where H 12 is obtained from P K 4ω,+ ω by adding an edge between two pendant vertices.

ω+2

1 From Lemma 2.2, we have q( H 12 ) > q( P K 4ω,+ ω ). If G − V ( K ω ) contains C 3 , note that G = P K 4,ω , then

ω+1 ω+1 G contains P K 3,ω as a proper subgraph. From Lemmas 2.2 and 2.4, we have q(G ) > q( P K 3, ω ) > ω+1 q( H 12 ) > q( P K 4,ω ). Case 2. There exists at least one vertex outside of K ω that is adjacent to at least two vertices 2 − v 1 (see Fig. 8) as a proper subgraph. From Lemmas 2.2, 2.4 and 3.12, of K ω . Then G contains M ω we have  2   ω +1    1 q(G ) > q M ω − v 1 > q( H 2 ) > q P K 3,ω > q( H 12 ) > q P K 4ω,+ ω .

Case 3. There exist at least two vertices outside of K ω that are adjacent to some vertices of K ω . Then G contains H 2 or H 3 as a proper subgraph. From Lemma 2.7 and the above inequalities, we 1 ω+1 have q( H 3 ) > q( H 2 ) > q( P K 4ω,+ ω ). From Lemma 2.2, we have q(G ) > q( P K 4,ω ). 2 Acknowledgement The authors would like to thank the anonymous referee for his valuable suggestions towards improving the paper and making the paper more readable.

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References [1] R. Merris, Laplacian graph eigenvectors, Linear Algebra Appl. 278 (1998) 221–236. [2] J.-M. Guo, J.X. Li, W.C. Shiu, The smallest Laplacian spectral radius of graphs with a given clique number, Linear Algebra Appl. 437 (2012) 1109–1122. ´ P. Hansen, The minimum spectral radius of graphs with a given clique number, Electron. J. Linear Algebra 17 [3] D. Stevanovic, (2008) 110–117. ´ P. Rowlinson, S.K. Simic, ´ A sharp lower bound for the least eigenvalue of the signless Laplacian [4] D.M. Cardoso, D. Cvetkovic, of a non-bipartite graph, Linear Algebra Appl. 429 (2008) 2770–2780. ´ P. Rowlinson, S.K. Simic, ´ Signless Laplacian of finite graphs, Linear Algebra Appl. 423 (2007) 155–171. [5] D. Cvetkovic, ´ S.K. Simic, ´ Towards a spectral theory of graphs based on the signless Laplacian I, Publ. Inst. Math. [6] D. Cvetkovic, (Beograd) 85 (99) (2009) 19–33. [7] J.M. Guo, J. Li, W.C. Shiu, On the Laplacian, signless Laplacian and normalized Laplacian characteristic polynomials of a graph, Czechoslovak Math. J. (2013), in press. [8] Y. Hong, X.D. Zhang, Sharp upper and lower bounds for largest eigenvalue of the Laplacian matrices of trees, Discrete Math. 296 (2005) 187–1973. ´ M. Doob, H. Sachs, Spectra of Graphs, third ed., Johann Ambrosius Barth Verlag, Heidelberg, Leipzig, 1995. [9] D. Cvetkovic, [10] B. Zhou, I. Gutman, A connection between ordinary and Laplacian spectra of bipartite graphs, Linear Multilinear Algebra 56 (2008) 305–310. [11] A.J. Schwenk, R.J. Wilson, On the eigenvalues of a graph, in: L.W. Beineke, R.J. Wilson (Eds.), Selected Topics in Graph Theory, Academic Press, London, 1978, pp. 307–336 (Chapter II). [12] A.J. Hoffman, J.H. Smith, On the spectral radii of topologically equivalent graphs, in: M. Fiedler (Ed.), Recent Advances in Graph Theory, Academic, Praha, 1975, pp. 273–281.